# silly question but on-topic!

• posted

Hello,

What happens to the charge in a polar liquid when it is no longer in the electric field that charged it? As in "flowing past charged plates".

Bill Martin

• posted

Wouldn't the energy simply be dissipated in rearranging the polar molecules back to a less "oraganised" state?

• posted

I would "guess" that to be the case, just curious if anyone ever looked into the situation more than casually.

Bill

• posted

Charging and then discharging a capacitor uses no net energy.

```--
John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com```
• posted

That is a simple corollary from the second law of thermodynamics: The kids' room gets messy by itself, and it takes plenty of energy to clean the mess.

```--

Tauno Voipio```
• posted

That's what made me curious. It seems like removing the charge from the system which delivered it may be a different situation...so it has to "go" somewhere in order to be conserved... consider the following simple picture:

+V | | ------------------- metal plate ----non-conducting tube wall---

--->flowing polar fluid --->

------------------- metal plate ----non-conducting tube wall--- | | | -V

Do you suppose it's possible to measure a current flow across the metal plates that relates to the bulk flow of the polar liquid through the "capacitor", if you will?

Bill

• posted

There would only be a steady-state current if power was being dissipated somewhere. The fluid is a dielectric insulator, a capacitor, so it won't get hot (unless it's a lossy capacitor). So the only way to lose power is by doing mechanical work.

I'm guessing that no work is done and no power is lost. There might be a pressure gradient caused by the field. There are probably mechanical forces on the liquid caused by the electric field, but I'm guessing they cancel.

Isn't a dielectric pulled into the field between two parallel plates? It's sort of like that.

So it should be possible to make an electrostatic pump.

```--
John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com```
• posted

Whether it's purely dielectric (displacement current only), resistive (lossy, conductive), ionic or other, this is the general subject which concerns it.

Note EHD mirrors MHD (magneto-...), and both in turn are closely associated with plasma physics (where E&M are both very important, because the working fluid is charged and aligns with electromagnetic fields, and flows at a rate where magnetic fields are significant).

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms```
• posted

when the field is no longer there brownian motion will stir them up.

```--
?? 100% natural

--- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net```
• posted

For a current to flow the liquid must accept electrons and "holes" from the plates I can't see a non-electrolyte doing that.

For an electrolyte the magnetohydrodynamic effect is probably simpler.

```--
?? 100% natural

--- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net```
• posted

Making a DC transformer?

-- Boris

• posted

If they're polar molecules that have been held in alignment by an external field then I'd have thought the polepole forces would far exceed Brownian forces at least at the beginning of the de-energisation of the electric field.

• posted

Not even wrong.

Charging or discharging a capacitor is typically less than 50% efficient, unless elaborate heroic measures are taken.

Take a charged 1 microfarad capacitor. Connect it to a uncharged 1 microfarad capacitor.

Voltage drops to half. Energy drops to one quarter.

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Many thanks,

Don Lancaster                          voice phone: (928)428-4073```
• posted

Heroic? One inductor will cyclically charge and discharge a capacitor with miniscule loss per cycle. Newbies like you should google "parallel resonant circuit."

Not even right.

```--
John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   ```
• posted

The two of you should be more specific. Don, emphasize asymptotic (infinite time), also a lossy circuit (though not even superconducting resonators are truely lossless (Q > 1e7) and will decay down to the noise floor over the span of perhaps minutes).

John, stop jumping on people with one arguably applicable statement when another was obviously implied (asymptotic vs. quarter cycle time in this case).

Speaking of which, some day I'm going to make a quasi-resonant gate drive and revolutionize the world on efficiency (if not necessarily physical size).

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms```
• posted

Isn't that derivable from, "climbing up and down a ladder does no work"?

Or does "uses no energy" simply mean that no energy was converted to another form?

```--
zero, and remove the last word.```
• posted

When people tell me that I'm "not even wrong", and then proceed to be wrong, I have a legal right to jump on them. Practically an obligation.

If you connect a battery to an uncharged cap, through an inductor and a diode, the cap gets charged to 2 * Vb, with a beautiful raised-cosine curve and 100% efficiency. That's the starting point.

```--
John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   ```
• posted

What part of "connect a cap to a cap" contains a diode? That's not the starting point.

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms```
• posted

Why do I get that déja-vu feeling?

```--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."```
• posted

You don't use diodes? Really, they can be handy.

```--
John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   ```

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